Trigonometry & Inverse Trigonometry
Trigonometry
star_batch_jee_advanced_2025
Grade 12
Question:
The solution(s) of the equation $\cos^{-1} x = \tan^{-1} x$ satisfy
$x^2 = \frac{\sqrt{5}-1}{2}$
$x^2 = \frac{\sqrt{5}+1}{2}$
$\sin(\cos^{-1} x) = \frac{\sqrt{5}-1}{2}$
$\tan(\cos^{-1} x) = \frac{\sqrt{5}-1}{2}$
Step-by-Step Solution
Key Concept: Setting $\cos^{-1}x = \tan^{-1}x$ and using the definitions of these inverse functions yields a polynomial equation in $x^2$ that reveals the relationship.
Let $\cos^{-1} x = \tan^{-1} x = \theta$. Then $x = \cos\theta$ and $\tan\theta = x$. From $\tan\theta = x$, we get $\sin\theta = \frac{x}{\sqrt{1+x^2}}$ and $\cos\theta = \frac{1}{\sqrt{1+x^2}}$. Since $x = \cos\theta$, we have $x = \frac{1}{\sqrt{1+x^2}}$. Squaring: $x^2(1+x^2) = 1$, giving $x^4 + x^2 - 1 = 0$. Using the quadratic formula: $x^2 = \frac{-1+\sqrt{5}}{2}$ (taking the positive root). For option 3, $\sin(\cos^{-1}x) = \sqrt{1-x^2} = \sqrt{1-\frac{\sqrt{5}-1}{2}} = \sqrt{\frac{\sqrt{5}-1}{2}}$, which after simplification equals $\frac{\sqrt{5}-1}{2}$.
Correct Answer: 1,3